Solve the equation.
4 cos² x = 1
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cos^2 x=1/4
cosx =1/2
cosx =cos 60°
x=2n pi + or-60°
cosx =1/2
cosx =cos 60°
x=2n pi + or-60°
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Solution of equation:

So there are two values of x can be calculated
![\cos(x) = \frac{1}{2} \\ \\ x = {cos}^{ - 1} ( \frac{1}{2} ) \\ \\ x = {cos}^{ - 1} (cos \: ( \frac{\pi}{3} ) \\ \\ = \frac{\pi}{3} \: \: belongs \: to \: [0,\pi] \cos(x) = \frac{1}{2} \\ \\ x = {cos}^{ - 1} ( \frac{1}{2} ) \\ \\ x = {cos}^{ - 1} (cos \: ( \frac{\pi}{3} ) \\ \\ = \frac{\pi}{3} \: \: belongs \: to \: [0,\pi]](https://tex.z-dn.net/?f=+%5Ccos%28x%29+%3D+%5Cfrac%7B1%7D%7B2%7D+%5C%5C+%5C%5C+x+%3D+%7Bcos%7D%5E%7B+-+1%7D+%28+%5Cfrac%7B1%7D%7B2%7D+%29+%5C%5C+%5C%5C+x+%3D+%7Bcos%7D%5E%7B+-+1%7D+%28cos+%5C%3A+%28+%5Cfrac%7B%5Cpi%7D%7B3%7D+%29+%5C%5C+%5C%5C+%3D+%5Cfrac%7B%5Cpi%7D%7B3%7D+%5C%3A+%5C%3A+belongs+%5C%3A+to+%5C%3A+%5B0%2C%5Cpi%5D)
![\cos(x) = - \frac{1}{2} \\ \\ x = {cos}^{ - 1} ( - \frac{1}{2} ) \\ \\ x = {cos}^{ - 1} ( - cos \: ( \frac{\pi}{3} ) \\ \\ = {cos}^{ - 1} (cos(\pi - \frac{\pi}{3} ) \\ \\ = {cos}^{ - 1}(cos( \frac{2\pi}{3})) \\ \\= \frac{2\pi}{3} \: \: belongs \: to \: [0,\pi]\\ \cos(x) = - \frac{1}{2} \\ \\ x = {cos}^{ - 1} ( - \frac{1}{2} ) \\ \\ x = {cos}^{ - 1} ( - cos \: ( \frac{\pi}{3} ) \\ \\ = {cos}^{ - 1} (cos(\pi - \frac{\pi}{3} ) \\ \\ = {cos}^{ - 1}(cos( \frac{2\pi}{3})) \\ \\= \frac{2\pi}{3} \: \: belongs \: to \: [0,\pi]\\](https://tex.z-dn.net/?f=%5Ccos%28x%29+%3D+-+%5Cfrac%7B1%7D%7B2%7D+%5C%5C+%5C%5C+x+%3D+%7Bcos%7D%5E%7B+-+1%7D+%28+-+%5Cfrac%7B1%7D%7B2%7D+%29+%5C%5C+%5C%5C+x+%3D+%7Bcos%7D%5E%7B+-+1%7D+%28+-+cos+%5C%3A+%28+%5Cfrac%7B%5Cpi%7D%7B3%7D+%29+%5C%5C+%5C%5C+%3D+%7Bcos%7D%5E%7B+-+1%7D+%28cos%28%5Cpi+-+%5Cfrac%7B%5Cpi%7D%7B3%7D+%29+%5C%5C+%5C%5C+%3D+%7Bcos%7D%5E%7B+-+1%7D%28cos%28+%5Cfrac%7B2%5Cpi%7D%7B3%7D%29%29+%5C%5C+%5C%5C%3D+%5Cfrac%7B2%5Cpi%7D%7B3%7D+%5C%3A+%5C%3A+belongs+%5C%3A+to+%5C%3A+%5B0%2C%5Cpi%5D%5C%5C)
Thus two values of x are π/3 and 2π/3.
So there are two values of x can be calculated
Thus two values of x are π/3 and 2π/3.
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